I was following this lecture on Newtons Laws.
https://youtu.be/2tHpgQmnH3A?si=Wbp36oBS_4b1HhIi
At 31:56 in the video, the board has a very general solution to Newton's second law. However the second component of the RHS of the final equation is a double integral, something I'm having a tough time grokking.
Why do the two integrals have different ranges? Aren't they being integrated over the same time period, lets say 0 to t, so why does the inner intergal have zero to t`?
We have
$$
v(t') =v(0)+ \int_0^{t'} a(t_1) dt_1=v(0)+\int_0^{t'}dt_1 F(t_1)/m.
$$
This follows from the fundamental theorem of calculus
$$
\frac{dv(t')}{dt'}= \frac{d}{dt'}\int_0^{t'} a(t_1) dt_1= a(t'),
$$
and the initial condition $v(t=0)=v(0)$. This inner integral goes only from $t=0$ to $t=t'$ becuase that is when we need $v(t)$ for the next integral:
$$
x(t)= \int_0^t v(t') dt'.
$$
Now just put them together to get
$$
x(t) = \int_0^t v(t') dt'= \int_0^t \left(v(0) +\int_0^{t'}dt_1 F(t_1)/m\right ) dt'\\
=v(0)t+ \int_0^t \left(\int_0^{t'}dt_1 F(t_1)/m\right ) dt'
$$
